Given that $4 \leq y \leq 6$ $4<=y<=6$ , find the value of y for which $2 cos (\frac{2 y}{3}) + \sqrt{3} = 0$ $2cos((2y)/3)+sqrt3=0$ ?

Question

1354 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-02-17T16:43:33

$y = \frac{5 π}{4}, \frac{7 π}{4}$ $color(blue)(y=(5pi)/4 , (7pi)/4)$

$2 cos (\frac{2 y}{3}) + \sqrt{3} = 0$ $2cos((2y)/3)+sqrt(3)=0$

$cos (\frac{2 y}{3}) = \frac{- \sqrt{3}}{2}$ $cos((2y)/3)=(-sqrt(3))/2$

Take arc cosine of both sides:

$arccos (cos (\frac{2 y}{3})) = arccos (- \frac{\sqrt{3}}{2})$ $arccos(cos((2y)/3))=arccos(-sqrt(3)/2)$

$\frac{2 y}{3} = arccos (- \frac{\sqrt{3}}{2})$ $(2y)/3=arccos(-sqrt(3)/2)$

$arccos (- \frac{\sqrt{3}}{2}) = \frac{5 π}{6}, \frac{7 π}{6}$ $arccos(-sqrt(3)/2)=(5pi)/6 , (7pi)/6$

We try these two first, to see if we are in the given interval.

$\frac{2 y}{3} = \frac{5 π}{6} \Rightarrow y = \frac{5 π}{4} \approx 3.927$ $(2y)/3=(5pi)/6=>y=(5pi)/4~~3.927$

$\frac{2 y}{3} = \frac{7 π}{6} \Rightarrow y = \frac{7 π}{4} \approx 5.498$ $(2y)/3=(7pi)/6=>y=(7pi)/4~~5.498$

Any greater angles will not be in the given interval.

So:

$y = \frac{5 π}{4}, \frac{7 π}{4}$ $color(blue)(y=(5pi)/4 , (7pi)/4)$

Given that 4≤y≤64<=y<=6, find the value of y for which 2cos(2y3)+√3=02cos((2y)/3)+sqrt3=0?